Integrated Rate Law Example 1

by Jules Bruno
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A plot of the concentration of nitrogen Tri Oxide vs time. With a slope of .260 gives a straight line. What was the initial concentration of nitrogen tri oxide? If after 35 seconds it's concentration dropped to 2.75 times 10 to the negative to moller. Alright, so here they're telling us it's a plot of concentration versus time. Remember if our plot is of your reacting concentration versus time? That would mean that it is a zero order reaction or zero order reaction. So we know it's zero order. So that means that our final reacting concentration equals negative Katie plus initial reacting concentration. Now here they're asking us to figure out the initial concentration. So we don't know what this portion is. We know that it drops to this final number here. So that's our final concentration. So that's 2.75 times 10 to the negative two equals. Now remember this equation. First order reaction is also equal to the equation for a straight line and here is our slope which is equal to our negative K. So when they tell me my slope is 0.260. They're really telling me what K. Is. So we're gonna plug that in for K. So negative 0.260 T. Here is time which we're told is 35 seconds. Alright, so then we're gonna have 2.75 times 10 to the negative two equals negative 9.1 plus the initial concentration of your reactant. Here, you're going to add 9.1 to both sides. And when we do that we're gonna get as our initial concentration 9.1275 moller. So this would represent our initial concentration here. In our question. This has three sig figs, three sig figs, two sig figs. So here, if we had it, in terms of two sig figs, it would come out to be uh 9.1 molar as our initial concentration, Right? So that would be our final answer.